1. **State the problem:** We need to evaluate the integral $$I = \int \frac{x^2(x^2+2)}{(x^2+1) (x \cos x - \sin x)^2} \, dx = \frac{f(x)}{g(x)} + C$$ and then identify functions $f(x)$ and $g(x)$ given conditions $f(0)=1$ and $g(0)=0$. 2. **Rewrite the denominator:** Note that $$x \cos x - \sin x = \sqrt{1+x^2} \left( \frac{x}{\sqrt{1+x^2}} \cos x - \frac{1}{\sqrt{1+x^2}} \sin x \right) = \sqrt{1+x^2} \cos(x + \alpha)$$ where $$\cos \alpha = \frac{x}{\sqrt{x^2+1}}, \quad \sin \alpha = \frac{1}{\sqrt{1+x^2}}, \quad \tan \alpha = \frac{1}{x}$$ 3. **Define substitution:** Let $$t = x + \tan^{-1}\left(\frac{1}{x}\right)$$ 4. **Calculate differential $dt$:** Differentiating, $$dt = 1 + \frac{d}{dx} \tan^{-1}\left(\frac{1}{x}\right) = 1 + \frac{1}{1+(1/x)^2} \times \left(-\frac{1}{x^2}\right) = 1 - \frac{1}{x^2+1} = \frac{x^2}{x^2+1}$$ 5. **Express the tangent of $t$:** Using tangent addition formula, $$\tan t = \frac{\tan x + \frac{1}{x}}{1 - \tan x \times \frac{1}{x}} = \frac{x \sin x + \cos x}{x \cos x - \sin x}$$ 6. **Express secant squared:** $$\sec^2 t = \left(\frac{1}{\cos t}\right)^2 = \frac{x^2 + 2}{(x \cos x - \sin x)^2}$$ 7. **Rewrite the integral with substitution:** Notice numerator can be expressed as: $$\frac{x^2(x^2+2)}{x^2+1} = (x^2 + 2) \times \frac{x^2}{x^2 + 1}$$ Substituting $dt$ from step 4, $$I = \int \sec^2 t \, dt$$ 8. **Integrate:** $$I = \tan t + C = \frac{x \sin x + \cos x}{x \cos x - \sin x} + C$$ 9. **Identify $f(x)$ and $g(x)$:** From the expression, $$\frac{f(x)}{g(x)} = \frac{x \sin x + \cos x}{x \cos x - \sin x}$$ so $$f(x) = x \sin x + \cos x, \quad g(x) = x \cos x - \sin x$$ Checking initial conditions, $$f(0) = 0 + 1 = 1, \quad g(0) = 0 - 0 = 0$$ which satisfy the problem's requirements. **Final Answer:** - (a) $f(x) = x \sin x + \cos x$ - (d) $g(x) = x \cos x - \sin x$